The equilibrium constant, ka, for the monoprotic acid is 7.08 × 10⁻¹⁶.
Acid dissociation constant (Ka) for monoprotic acid will be calculated according to the given equation:
Ka = [H₃O⁺][A⁻] / [HA]
Relation between H₃O⁺ and pH will be represented as:
[H₃O⁺] = [tex]10^{-pH}[/tex]
pH = 6.44 (given)
[H₃O⁺] = [tex]10^{-6.44}[/tex] = 3.63×10⁻⁷
Given chemical reaction with ICE table is:
HA(aq) + H₂O(l) ⇄ H₃O⁺(aq) + A⁻
Initial: 0.0186 0 0
Change: -3.63×10⁻⁷ 3.63×10⁻⁷ 3.63×10⁻⁷
Equilibrium: 0.0186-3.63×10⁻⁷ 3.63×10⁻⁷ 3.63×10⁻⁷
Now Ka for this will be:
Ka = (3.63×10⁻⁷)² / (0.0186-3.63×10⁻⁷)
Value of 3.63×10⁻⁷ is negligible as compared to the 0.0186 value and equation becomes,
Ka = (3.63×10⁻⁷)² / 0.0186 = 13.18 × 10⁻¹⁴ / 0.0186
Ka = 7.08 × 10⁻¹⁶
Hence value of equilibrium constant is 7.08 × 10⁻¹⁶.
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